I have a pandas dataframe with one column that contains an empty list in each cell.
I need to duplicate the dataframe, and append it at the bottom of the original dataframe, but with additional information in the list.
Here is a minimal code example:
df_main = pd.DataFrame([['a', []], ['b', []]], columns=['letter', 'mylist'])
> df_main
letter mylist
0 a []
1 b []
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = row.mylist.append(1)
pd.concat([ df_copy,df_main], ignore_index=True)
> result:
letter mylist
0 a None
1 b None
2 a [1]
3 b [1]
As you can see there is a problem that the [] empty list was replaced by a None
Just to make sure, this is what I would like to have:
letter mylist
0 a []
1 b []
2 a [1]
3 b [1]
How can I achieve that?
append method on list return a None value, that's why None appears in the final dataframe. You may have use + operator for reassignment like this:
import pandas as pd
df_main = pd.DataFrame([['a', []], ['b', []]], columns=['letter', 'mylist'])
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = row.mylist + list([1])
pd.concat([df_main, df_copy], ignore_index=True).head()
Output of this block of code:
letter mylist
0 a []
1 b []
2 a [1]
3 b [1]
A workaround to solve your problem would be to create a temporary column mylist2 with np.empty((len(df), 0)).tolist()) and use np.where() to change the None values of mylist to an empty list and then drop the empty column.
import pandas as pd, numpy as np
df_main = pd.DataFrame([['a', []], ['b', []]], columns=['letter', 'mylist'])
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = row.mylist.append(1)
df = (pd.concat([df_copy,df_main], ignore_index=True)
.assign(mylist2=np.empty((len(df), 0)).tolist()))
df['mylist'] = np.where((df['mylist'].isnull()), df['mylist2'], df['mylist'])
df= df.drop('mylist2', axis=1)
df
Out[1]:
letter mylist
0 a []
1 b []
2 a [1]
3 b [1]
Not only does append method on list return a None value as indicated in the first answer, but both df_main and df_copy contain pointers to the same lists. So after:
for index, row in df_copy.iterrows():
row.mylist.append(1)
both dataframes have updated lists with one element. For your code to work as expected you can create a new list after you copy the dataframe:
df_copy = df_main.copy()
for index, row in df_copy.iterrows():
row.mylist = []
This question is another great example why we should not put objects in a dataframe.
Related
import numpy as np
import pandas as pd
d = {'ABSTRACT_ID': [14145090,1900667, 8157202,6784974],
'TEXT': [
"velvet antlers vas are commonly used in tradit",
"we have taken a basic biologic RPA to elucidat4",
"ceftobiprole bpr is an investigational cephalo",
"lipoperoxidationderived aldehydes for example",],
'LOCATION': [1, 4, 2, 1]}
df = pd.DataFrame(data=d)
df
def word_at_pos(x,y):
pos=x
string= y
count = 0
res = ""
for word in string:
if word == ' ':
count = count + 1
if count == pos:
break
res = ""
else :
res = res + word
print(res)
word_at_pos(df.iloc[0,2],df.iloc[0,1])
For this df I want to create a new column WORD that contains the word from TEXT at the position indicated by LOCATION. e.g. first line would be "velvet".
I can do this for a single line as an isolated function world_at_pos(x,y), but can't work out how to apply this to whole column. I have done new columns with Lambda functions before, but can't work out how to fit this function to lambda.
Looping over TEXT and LOCATION could be the best idea because splitting creates a jagged array, so filtering using numpy advanced indexing won't be possible.
df["WORDS"] = [txt.split()[loc] for txt, loc in zip(df["TEXT"], df["LOCATION"]-1)]
print(df)
ABSTRACT_ID ... WORDS
0 14145090 ... velvet
1 1900667 ... a
2 8157202 ... bpr
3 6784974 ... lipoperoxidationderived
[4 rows x 4 columns]
I can do the following if I want to extract rows whose column "A" contains the substring "hello".
df[df['A'].str.contains("hello")]
How can I select rows whose column is the substring for another word? e.g.
df["hello".contains(df['A'].str)]
Here's an example dataframe
df = pd.DataFrame.from_dict({"A":["hel"]})
df["hello".contains(df['A'].str)]
IIUC, you could apply str.find:
import pandas as pd
df = pd.DataFrame(['hell', 'world', 'hello'], columns=['A'])
res = df[df['A'].apply("hello".find).ne(-1)]
print(res)
Output
A
0 hell
2 hello
As an alternative use __contains__
res = df[df['A'].apply("hello".__contains__)]
print(res)
Output
A
0 hell
2 hello
Or simply:
res = df[df['A'].apply(lambda x: x in "hello")]
print(res)
I am trying to apply a function to every column in a dataframe, when I try to do it on just a single fixed column name it works. I tried doing it on every column, but when I try passing the column name as an argument in the function I get an error.
How do you properly pass arguments to apply a function on a data frame?
def result(row,c):
if row[c] >=0 and row[c] <=1:
return 'c'
elif row[c] >1 and row[c] <=2:
return 'b'
else:
return 'a'
cols = list(df.columns.values)
for c in cols
df[c] = df.apply(result, args = (c), axis=1)
TypeError: ('result() takes exactly 2 arguments (21 given)', u'occurred at index 0')
Input data frame format:
d = {'c1': [1, 2, 1, 0], 'c2': [3, 0, 1, 2]}
df = pd.DataFrame(data=d)
df
c1 c2
0 1 3
1 2 0
2 1 1
3 0 2
You don't need to pass the column name to apply. As you only want to check if values of the columns are in certain range and should return a, b or c. You can make the following changes.
def result(val):
if 0<=val<=1:
return 'c'
elif 1<val<=2:
return 'b'
return 'a'
cols = list(df.columns.values)
for c in cols
df[c] = df[c].apply(result)
Note that this will replace your column values.
A faster way is np.select:
import numpy as np
values = ['c', 'b']
for col in df.columns:
df[col] = np.select([0<=df[col]<=1, 1<df[col]<=2], values, default = 'a')
Blockquote
I have the following dataframe.
d = pd.DataFrame({'a': [['foo', 'bar'], ['bar'], ['fah', 'baz']})
I'd like to return just the rows with values of a beginning f in them - i.e. the first and third rows.
This is what I've tried:
d[d.a.is_in('f')]
Use any in list comprehension with generator:
d = d[[any(y.startswith('f') for y in x) for x in d['a']]]
print (d)
a
0 [foo, bar]
2 [fah, baz]
Detail: (convert to list only for sample)
print ([list(y.startswith('f') for y in x) for x in d['a']])
[[True, False], [False], [True, False]]
Solution using .apply(), iterating over the individual list elements, checking with .startswith() and evaluating the length of the resultant list:
import pandas as pd
df = pd.DataFrame({'a': [['foo', 'bar'], ['bar'], ['fah', 'baz']]})
df = df[df.a.apply(lambda x: len([el for el in x if el.startswith('f')]) > 0)]
print(df)
which results in:
a
0 [foo, bar]
2 [fah, baz]
I have a dataframe with column names, and I want to find the one that contains a certain string, but does not exactly match it. I'm searching for 'spike' in column names like 'spike-2', 'hey spike', 'spiked-in' (the 'spike' part is always continuous).
I want the column name to be returned as a string or a variable, so I access the column later with df['name'] or df[name] as normal. I've tried to find ways to do this, to no avail. Any tips?
Just iterate over DataFrame.columns, now this is an example in which you will end up with a list of column names that match:
import pandas as pd
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)
spike_cols = [col for col in df.columns if 'spike' in col]
print(list(df.columns))
print(spike_cols)
Output:
['hey spke', 'no', 'spike-2', 'spiked-in']
['spike-2', 'spiked-in']
Explanation:
df.columns returns a list of column names
[col for col in df.columns if 'spike' in col] iterates over the list df.columns with the variable col and adds it to the resulting list if col contains 'spike'. This syntax is list comprehension.
If you only want the resulting data set with the columns that match you can do this:
df2 = df.filter(regex='spike')
print(df2)
Output:
spike-2 spiked-in
0 1 7
1 2 8
2 3 9
This answer uses the DataFrame.filter method to do this without list comprehension:
import pandas as pd
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6]}
df = pd.DataFrame(data)
print(df.filter(like='spike').columns)
Will output just 'spike-2'. You can also use regex, as some people suggested in comments above:
print(df.filter(regex='spike|spke').columns)
Will output both columns: ['spike-2', 'hey spke']
You can also use df.columns[df.columns.str.contains(pat = 'spike')]
data = {'spike-2': [1,2,3], 'hey spke': [4,5,6], 'spiked-in': [7,8,9], 'no': [10,11,12]}
df = pd.DataFrame(data)
colNames = df.columns[df.columns.str.contains(pat = 'spike')]
print(colNames)
This will output the column names: 'spike-2', 'spiked-in'
More about pandas.Series.str.contains.
# select columns containing 'spike'
df.filter(like='spike', axis=1)
You can also select by name, regular expression. Refer to: pandas.DataFrame.filter
df.loc[:,df.columns.str.contains("spike")]
Another solution that returns a subset of the df with the desired columns:
df[df.columns[df.columns.str.contains("spike|spke")]]
You also can use this code:
spike_cols =[x for x in df.columns[df.columns.str.contains('spike')]]
Getting name and subsetting based on Start, Contains, and Ends:
# from: https://stackoverflow.com/questions/21285380/find-column-whose-name-contains-a-specific-string
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.Series.str.contains.html
# from: https://cmdlinetips.com/2019/04/how-to-select-columns-using-prefix-suffix-of-column-names-in-pandas/
# from: https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.filter.html
import pandas as pd
data = {'spike_starts': [1,2,3], 'ends_spike_starts': [4,5,6], 'ends_spike': [7,8,9], 'not': [10,11,12]}
df = pd.DataFrame(data)
print("\n")
print("----------------------------------------")
colNames_contains = df.columns[df.columns.str.contains(pat = 'spike')].tolist()
print("Contains")
print(colNames_contains)
print("\n")
print("----------------------------------------")
colNames_starts = df.columns[df.columns.str.contains(pat = '^spike')].tolist()
print("Starts")
print(colNames_starts)
print("\n")
print("----------------------------------------")
colNames_ends = df.columns[df.columns.str.contains(pat = 'spike$')].tolist()
print("Ends")
print(colNames_ends)
print("\n")
print("----------------------------------------")
df_subset_start = df.filter(regex='^spike',axis=1)
print("Starts")
print(df_subset_start)
print("\n")
print("----------------------------------------")
df_subset_contains = df.filter(regex='spike',axis=1)
print("Contains")
print(df_subset_contains)
print("\n")
print("----------------------------------------")
df_subset_ends = df.filter(regex='spike$',axis=1)
print("Ends")
print(df_subset_ends)